I need this question\'s Answer on Microsoft Excel. THank You. A courier service

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I need this question's Answer on Microsoft Excel. THank You.

A courier service advertises that its average delivery time is less than six hours for local deliveries. A random sample of the amount of time this courier takes to deliver packages to an address across town produced the following times (rounded to the nearest hour): 7,3,4, 6,10,5, 6, 4, 3, and 8. Is this sufficient evidence to support the courier's advertisement, at the 5 percent level of significance? Find tire 99 percent confidence interval estimate of mean delivery time. What assumption must be made in order to answer these questions?

Explanation / Answer

1.
Set Up Hypothesis
Null, H0: U=6
Alternate, Average delivery time is less than 6 Hours, H1: U<6
Test Statistic
Population Mean(U)=6
Sample X(Mean)=5.6
Standard Deviation(S.D)=2.271
Number (n)=10
we use Test Statistic (t) = x-U/(s.d/Sqrt(n))
to =5.6-6/(2.271/Sqrt(9))
to =-0.557
| to | =0.557
Critical Value
The Value of |t | with n-1 = 9 d.f is 1.833
We got |to| =0.557 & | t | =1.833
Make Decision
Hence Value of |to | < | t | and Here we Do not Reject Ho
P-Value :Left Tail -Ha : ( P < -0.557 ) = 0.29556
Hence Value of P0.05 < 0.29556,Here We Do not Reject Ho

We don't have the evidence to indicate that Average delivery time is less than 6 Hours

2.
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x)=5.6
Standard deviation( sd )=2.271
Sample Size(n)=10
Confidence Interval = [ 5.6 ± t a/2 ( 2.271/ Sqrt ( 10) ) ]
= [ 5.6 - 3.25 * (0.718) , 5.6 + 3.25 * (0.718) ]
= [ 3.266,7.934 ]

3.
Sample is under normal

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